Find the equations of the two lines passing through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi}{3}$ each.

(A) Let the equation of the given line be $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\lambda$.
Any point on this line is $P(2\lambda+3, \lambda+3, \lambda)$.
The direction ratios of the line are $(2, 1, 1)$.
Since the required lines pass through the origin $(0, 0, 0)$,the direction ratios of the required lines are $(2\lambda+3, \lambda+3, \lambda)$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Given $\theta = \frac{\pi}{3}$,so $\cos \frac{\pi}{3} = \frac{1}{2}$.
Substituting the values:
$\frac{1}{2} = \frac{|2(2\lambda+3) + 1(\lambda+3) + 1(\lambda)|}{\sqrt{2^2+1^2+1^2} \sqrt{(2\lambda+3)^2 + (\lambda+3)^2 + \lambda^2}}$
$\frac{1}{2} = \frac{|4\lambda+6+\lambda+3+\lambda|}{\sqrt{6} \sqrt{4\lambda^2+12\lambda+9 + \lambda^2+6\lambda+9 + \lambda^2}}$
$\frac{1}{2} = \frac{|6\lambda+9|}{\sqrt{6} \sqrt{6\lambda^2+18\lambda+18}}$
Squaring both sides:
$\frac{1}{4} = \frac{(6\lambda+9)^2}{6(6\lambda^2+18\lambda+18)}$
$6(6\lambda^2+18\lambda+18) = 4(36\lambda^2+108\lambda+81)$
$36\lambda^2+108\lambda+108 = 144\lambda^2+432\lambda+324$
$108\lambda^2+324\lambda+216 = 0$
Dividing by $108$:
$\lambda^2+3\lambda+2 = 0 \Rightarrow (\lambda+1)(\lambda+2) = 0$.
So,$\lambda = -1$ or $\lambda = -2$.
For $\lambda = -1$,the direction ratios are $(2(-1)+3, -1+3, -1) = (1, 2, -1)$.
For $\lambda = -2$,the direction ratios are $(2(-2)+3, -2+3, -2) = (-1, 1, -2)$.
Thus,the equations of the lines are $\frac{x}{1} = \frac{y}{2} = \frac{z}{-1}$ and $\frac{x}{-1} = \frac{y}{1} = \frac{z}{-2}$.